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How many $4$-digit numbers, each greater than $1000$ and each having all four digits distinct, are there with $7$ coming before $3.$

The $4$-digit numbers can be from such that each is greater than $1000.$ And $7$ coming before $3.$

$\textbf{Case 1:}$

• $\underline{\boxed{7}} \quad \underline{{\color{Red} {3}}} \quad \underbrace{\underline{}}_{8\; \text{ways}} \quad \underbrace{\underline{}}_{7\; \text{ways}} = 8 \times 7 = 56$
• $\underline{\boxed{7}} \quad \underbrace{\underline{}}_{8\; \text{ways}} \quad \underline{{\color{Red} {3}}} \quad \underbrace{\underline{}}_{7\; \text{ways}} = 8 \times 7 = 56$
• $\underline{\boxed{7}} \quad \underbrace{\underline{}}_{8\; \text{ways}} \quad \underbrace{\underline{}}_{7\; \text{ways}} \quad \underline{{\color{Red} {3}}} = 8 \times 7 = 56$

The number of ways $= 3 \times 56 = 168 \; \text{ways.}$

$\textbf{Case 2:}$

• $\underbrace{\underline{}}_{7\; \text{ways}} \quad \underline{\boxed{7}} \quad \underline{{\color{Blue} {3}}} \quad \underbrace{\underline{}}_{7\; \text{ways}} = 7 \times 7 = 49$
• $\underbrace{\underline{}}_{7\; \text{ways}} \quad \underline{\boxed{7}} \quad \underbrace{\underline{}}_{7\; \text{ways}} \quad \underline{{\color{Blue} {3}}} = 7 \times 7 = 49$

The number of ways $= 2 \times 49 = 98 \; \text{ways}.$

$\textbf{Case 3:}$

• $\underbrace{\underline{}}_{7\; \text{ways}} \quad \underbrace{\underline{}}_{7\; \text{ways}} \underline{\boxed{7}} \quad \underline{{\color{Magenta} {3}}} = 7 \times 7 = 49$

The number of ways $= 49 \; \text{ways}.$

Thus, total such four digit numbers $= 168 + 98 + 49 = 315.$

$\therefore$ The $4 – \text{digit}$ number greater than $1000$ with $7$ before $3$ is $315.$

Correct Answer$: 315$

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